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To show: $13\, |\, 4^{2n+1}+3^{n+2}$

I used induction beginning successfully with $n=0$ $($or $n=1),$ then making the step to $n+1:$

An $x$ exists so that $13x = 4^{2n+3}+3^{n+3}$

$$\begin{align*}13x &= 16\cdot4^{2n+1}+3\cdot3^{n+2}\\ \frac{13}{3}x &= \frac{16}{3}\cdot4^{2n+1}+3^{n+3}\\ \frac{13}{3}x &= \frac{13}{3}\cdot4^{2n+1}+4^{2n+1}+3^{n+3}\end{align*}$$

Here comes the actual question: Can I say that because of the premise the last two summands can be expressed as $13m$ with $m$ being a natural number? Like this:

$$\begin{align*}\frac{13}{3}x &= \frac{13}{3}\cdot4^{2n+1}+13m\\ x &= 4^{2n+1}+3m,\end{align*}$$ which is a natural number.

Is that a valid proof?

355durch113
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    You should write: an integer $x$ exists such that $13x = \ldots \iff 13x/3 = \ldots \iff x = \ldots,$ because you need the $\Leftarrow$'s to conclude that $,x,$ is a solution of the original equation, i.e. in your proof, you need to go from bottom to top. – Bill Dubuque Mar 24 '14 at 01:14

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Yes, it seems to be correct. However, the person who would read the proof might be confused a little. Instead, you can re-write it differently, like $$ \frac{4^{2n+3} + 3^{n+3}}{13} = \frac{13 \cdot 4^{2n+1} + 3\left(4^{2n+1} + 3^{n+1} \right)}{13} = \frac{13\cdot 4^{2n+1} + 3 \cdot 13m}{13} \in \mathbb{Z} $$

which is the same, except it's going from 'bottom up'.

Husain
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